An algebra problem by rene lopez

x-y=0

so,

x=y

x+y=20

2x=20

x=10,y=10

xy=10X10=100

$\boxed {x + y = 20}$

$\boxed {x - y = 0}$

This shows that $x$ and $y$ must be positive and equal

$(x + y) + (x - y) = (20) + (0)$

$2x = 20$

$x = 10$

By algebra, $y = 10$

If $y = 10$ and $x = 10$ ; then

$xy = 10 × 10$

$xy = 100$

Firstly (x+y)+(x-y)=20 x+x+y-y=20
2x=20
x=10 then we have to find y
x+y=20
10+y=20
y=20-10
y=10
therefore xy = 10×10
= 100

xy=10*10=100 xy=100;

solving the 2 small equations we get 2x=20, x=10 and now in first equation y=20-10 = 10...so xy=10 x 10=100

Notice that $x=y$ from the second information, and $x+x=20$ from the first information, $x=10$ . $xy=xx=x^2=10^2=100$ .

Note that $(x+y)+(x-y)=2x=20 \implies x=10$ . Substituting this into the first equation, we have $y=10$ and hence $xy=\boxed{100}$ .